Applying a voltage to a homogeneous semiconductor

Question

Imagine that there is a homogeneous semiconductor (either a p-type or an n-type, but just one of them) uniformly doped. There is no diffusion current because of the uniform dopage, and there is no drift current because there is no electric field, as the charge is the same everywhere.

But what would happen if I applied a voltage to it? There would be an electric field across the semiconductor, so I guess that there would be a drift current. However, this would lead to carriers moving to one side of the semiconductor, and so there would be now a difference in concentration that would lead to difussion current. However, this doesn't sound right to me. What would actually happen?

Answer 1

Your scenario assumes that you are not able to inject/extract electrons into your semiconductor at wherever the contacts exist. In other words, you have infinite contact resistance, and in that case, yes, electrons and holes would build up on opposite sides. All voltage drop would occur at these charge buildup regions at the edges of the semiconductor, and the Fermi level is flat in between. It's the same as a metal floating in an electric field but not contacted; the charges move to cancel out the electric field on the inside.

In the opposite extreme, if the contacts are perfect, all the voltage drop will occur in the semiconductor. You'd have a linear sloped Fermi level, and bands parallel to that. There would be a drift, but not diffusion, current.

The stuff about photoconductivity above is extraneous.

Answer 2

A chunk of semiconductor would have some resistivity, and at low field it's just another resistor. If the field is high enough, the charge carriers will avalanche, though, and the resistivity will go WAY down until/unless the field drops (and there will be a delay while the free charges return to equilibrium).

The resistivity will also go down if you illuminate the semiconductor (i.e. it is photoconductive). Again, the resistivity doesn't return to its dark value immediately.

Photoconductors and some kinds of breakover devices (MOVs yes, diacs no) and many strain gages are just electrodes and a uniformly doped semiconductor.

The semiconductor's (low-current) resistivity temperature coefficient is negative (higher temperature causes resistance to drop), which is the opposite of resistivity temperature coefficient of a metal.

Answer 3

Most of possible scenarios and associated behaviors have already been covered in other answers and comments. Still, some clarification could potentially be useful.

What happens depends on the level of doping.

If the level of doping is high, the contacts between the semiconductor and the external circuit (presumably involving metal wires) will be ohmic and the applied voltage will cause a current to flow. This would be a drift current, i.e., a current due to the applied electric field.

In this case, the semiconductor will behave as a small resistor with a little twist: due to the Peltier effect, one of the contacts will be cooled and the other heated.

If the level of doping is low, the contacts will be rectifying (Schottky contacts) and, since one of them will always be reverse biased, there won't be much of a current flow. All the applied voltage will drop across the contact that happens to be reverse biased. From the outside it'll look just as a capacitor with a small leakage.

After the initial current, charging the capacitance of the reverse biased contact, there won't be any measurable voltage drop or electric field across the bulk of the semiconductor and there won't be a gradient of charges and associated diffusion and drift currents.

The scenario you are describing, when the electrons get redistributed across the bulk of the semiconductor, could occur if we apply an external electrostatic field without making electrical contacts with the semiconductor (the second case outlined by Jon Custer).

In this scenario, the electrons will redistribute themselves to equalize the potential across the semiconductor and we can say that there would be a dynamic equilibrium between the diffusion and drift currents with the net current of zero.

Answer 4

Consider following semiconductor bar with applied external voltage source. The main factor of current is drift current.

In order to consider diffusion current you should have gradient of minority carriers. However, in this situation the charge enter to unit volume is equal to the charge leave to unit volume since there is no charge build up in semiconductor bar and the bar is uniformly doped.